What is the $p H$ of a $0.200$ M solution of $C_{6} H_{5} C O O N a$ ?
(The $K_{a}$ of $C_{6} H_{5} C O O H$ is $6.4 \times 10^{- 5}$ )

5.25

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5.40

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8.60

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8.75

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Solution

The correct option is A $8.75$
$C_{6} H_{5} C O O N a$ is a salt made from weak acid $C_{6} H_{5} C O O H$ and strong base $N a O H$ . $p H$ will be determined by the hydrolysis of $C_{6} H_{5} C O O N a$ .
$C_{6} H_{5} C O O N a + H_{2} O ⇌ C_{6} H_{5} C O O H + N a^{+} + O H^{⊝}$
$∴ p H = \frac{1}{2} p K_{w} + \frac{1}{2} p K_{a} + \frac{1}{2} log C$
$= \frac{1}{2} \times 14 + \frac{1}{2} (- log K_{a}) + \frac{1}{2} log 0.2$
$= 7 + \frac{1}{2} \times (- log 6.4 \times 10^{- 5}) + \frac{1}{2} log 0.2$
$= 8.747 \approx 8.75$

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